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How Does Operator Overloading Work with Member Access Operators (->, .)?

Patricia Arquette
Release: 2024-12-03 07:28:10
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How Does Operator Overloading Work with Member Access Operators (->, .)?
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Understanding Member Access Operator Overloading

While operator overloading is generally straightforward, understanding the member access operators ->, ., -> etc. can be confusing. This article delves into their nuanced behavior and provides answers to common questions.

Passed Arguments and Returned Values

The operator-> function takes no arguments and returns the object it is called on. This allows for subsequent member lookups to be handled by another operator-> call. The operator-* function takes and returns whatever arguments you specify.

Identifying Referenced Members

The operator-> function does not need to know the specific member being referenced as it simply returns the containing object. However, the subsequent member lookup does use the return value.

const Considerations

Overloading operator-> requires a non-const version. Overloading operator-> does not have specific const considerations. Operators . and . cannot be overloaded.

Sample Code

Consider the following example demonstrating operator->:

struct client { int a; };

struct proxy {
    client *target;
    client *operator->() const { return target; }
};

void f() {
    client x{3};
    proxy y{&x};
    std::cout << x.a << y->a;    // prints "33"
}
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This code showcases the "drill-down" behavior of operator->, where multiple operator-> calls are chained together to perform member lookups.

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